The polynomial $p(x)=2x^3+17x^2+41x+30$ has a known factor of $(x+5)$. Rewrite $p(x)$ as a product of linear factors. $p(x)=$
Explanation: We know $(x+5)$ is a factor of $p(x)$. This means that $p(x)=(x+5)\cdot q(x)$ for some polynomial $q(x)$. We can find $q(x)$ using polynomial division, and then we can factor $q(x)$. This way, we will be able to rewrite $p(x)$ as a product of linear factors. Dividing $p(x)$ by $(x+5)$ $\begin{array}{r} 2x^2+\phantom{1}7x+\phantom{1}6 \\ x+5|\overline{2x^3+17x^2+41x+30} \\ \mathllap{-(}\underline{2x^3+10x^2\phantom{+41x+30}\rlap )} \\ 7x^2+41x+30 \\ \mathllap{-(}\underline{7x^2+35x\phantom{+30}\rlap )} \\ 6x+30 \\ \mathllap{-(}\underline{6x+30\rlap )} \\ 0 \end{array}$ We find that $q(x)=2x^2+7x+6$. Factoring $q(x)$ We can factor $q(x)$ by grouping: $\begin{aligned} q(x)&=2x^2+7x+6 \\\\ &=2x^2+4x+3x+6 \\\\ &=2x(x+2)+3(x+2) \\\\ &=(2x+3)(x+2) \end{aligned}$ Putting it all together $\begin{aligned} p(x)&=2x^3+17x^2+41x+30 \\\\ &=(x+5)(2x^2+7x+6) \\\\ &=(x+5)(2x+3)(x+2) \end{aligned}$